# Sum of Functions

Let $$f(x)=x-\dfrac{1}{x}$$.

(a) If $$f(x)+f(y)=0$$, then prove or disprove that $$f(x^n)+f(y^n)=0$$ for all integers $$n$$.

(b) If $$f(x)+f(y)+f(z)=0$$, then prove or disprove that $$f(x^n)+f(y^n)+f(z^n)=0$$ for all integers $$n$$.

(c) Do the above two problems with the added restriction that $$x,y,z$$ are positive.

Note by Daniel Liu
3 years, 11 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

(a) Counterexample: $$y=-x$$ and $$n=2a$$ for some positive integer $$a$$. $$f(x)+f(-x)=x-\dfrac{1}{x}-x+\dfrac{1}{x}=0$$, but $$f(x^n)+f((-x)^n)=f(x^{2a})+f(x^{2a})=x^{2a}-\dfrac{1}{x^{2a}}+x^{2a}-\dfrac{1}{x^{2a}}\neq0$$.

(b) Again, the counterexample is $$y=-x$$, $$z=1$$, and $$n=2a$$ for some positive integer $$a$$. $$f(x)+f(-x)+f(1)=x-\frac{1}{x}-x+\frac{1}{x}+1-1=0$$, but $$f(x^n)+f((-x)^n)+f(1^n)=f(x^{2a})+f(x^{2a})=f(1)=x^{2a}-\dfrac{1}{x^{2a}}+x^{2a}-\dfrac{1}{x^{2a}}+1-1\neq0$$.

(c) (part 1): Since $$x,y$$ are positive and $$f(x)+f(y)=0$$, $$x-\dfrac{1}{x}+y-\dfrac{1}{y}=0$$, and $$x+y=\dfrac{1}{x}+\dfrac{1}{y}$$. Adding the fractions on the RHS gives $$x+y=\dfrac{x+y}{xy}$$ so $$1=\dfrac{1}{xy}$$ or $$xy=1$$. Therefore, $$f(x^n)+f(y^n)=x^n-\dfrac{1}{x^n}+y^n-\dfrac{1}{y^n}=x^n-\dfrac{1}{x^n}+\dfrac{1}{x^n}-x^n=0$$.

- 3 years, 11 months ago

Sorry, your answer for part (a) is wrong. Also, the restriction of part (c) actually matters.

EDIT: thanks, fixed!

- 3 years, 11 months ago